poj 2976: Dropping tests(01分数规划--Dinkelbach算法)
2017-08-23 17:50
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Dropping tests
Description
In a certain course, you take n tests. If you get ai out of bi questions correct on test i, your cumulative average is defined to be
.
Given your test scores and a positive integer k, determine how high you can make your cumulative average if you are allowed to drop any k of your test scores.
Suppose you take 3 tests with scores of 5/5, 0/1, and 2/6. Without dropping any tests, your cumulative average is
. However, if you drop the third test, your cumulative average becomes
.
Input
The input test file will contain multiple test cases, each containing exactly three lines. The first line contains two integers, 1 ≤ n ≤ 1000 and 0 ≤ k < n. The second line contains n integers indicating ai for
all i. The third line contains n positive integers indicating bi for all i. It is guaranteed that 0 ≤ ai ≤ bi ≤ 1, 000, 000, 000. The end-of-file is marked by a test case
with n = k = 0 and should not be processed.
Output
For each test case, write a single line with the highest cumulative average possible after dropping k of the given test scores. The average should be rounded to the nearest integer.
Sample Input
Sample Output
题意:
给你n组数,每组数中有两个数a[i]和b[i],让你在这n组数中选出n-k组使得100*∑a[i]/∑b[i]值最大
经典01分数规划问题:
http://blog.csdn.net/jaihk662/article/details/77505318
还是这个公式,x[i]表示选还是不选
先给出二分思路:
二分cnt,然后求出所有的d[i],按d[i]从大到小排序选择前n-k个并求和,如果大于0说明可能不是最优解
再说Dinkelbach算法:
先随机一个cnt,然后求出所有的d[i],按d[i]从大到小排序选择前n-k个求出∑a[i]/∑b[i],如果∑a[i]/∑b[i]==cnt,那么说明是最优解,否则令cnt=∑a[i]/∑b[i]继续,直到求出最优解(一般优于二分)
其实这题还有一个贪心思路:算出所有的d[i] = a[i]/b[i],也就是比率,然后按这个从大到小排序,选最大的k个
但这是错的,一个很简单的反例:
3 1
100 1 100
100 5 200
正确答案应该是(101/105)*100 ≈ 96
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 13323 | Accepted: 4675 |
In a certain course, you take n tests. If you get ai out of bi questions correct on test i, your cumulative average is defined to be
.
Given your test scores and a positive integer k, determine how high you can make your cumulative average if you are allowed to drop any k of your test scores.
Suppose you take 3 tests with scores of 5/5, 0/1, and 2/6. Without dropping any tests, your cumulative average is
. However, if you drop the third test, your cumulative average becomes
.
Input
The input test file will contain multiple test cases, each containing exactly three lines. The first line contains two integers, 1 ≤ n ≤ 1000 and 0 ≤ k < n. The second line contains n integers indicating ai for
all i. The third line contains n positive integers indicating bi for all i. It is guaranteed that 0 ≤ ai ≤ bi ≤ 1, 000, 000, 000. The end-of-file is marked by a test case
with n = k = 0 and should not be processed.
Output
For each test case, write a single line with the highest cumulative average possible after dropping k of the given test scores. The average should be rounded to the nearest integer.
Sample Input
3 1 5 0 2 5 1 6 4 2 1 2 7 9 5 6 7 9 0 0
Sample Output
83 100
题意:
给你n组数,每组数中有两个数a[i]和b[i],让你在这n组数中选出n-k组使得100*∑a[i]/∑b[i]值最大
经典01分数规划问题:
http://blog.csdn.net/jaihk662/article/details/77505318
还是这个公式,x[i]表示选还是不选
先给出二分思路:
二分cnt,然后求出所有的d[i],按d[i]从大到小排序选择前n-k个并求和,如果大于0说明可能不是最优解
再说Dinkelbach算法:
先随机一个cnt,然后求出所有的d[i],按d[i]从大到小排序选择前n-k个求出∑a[i]/∑b[i],如果∑a[i]/∑b[i]==cnt,那么说明是最优解,否则令cnt=∑a[i]/∑b[i]继续,直到求出最优解(一般优于二分)
其实这题还有一个贪心思路:算出所有的d[i] = a[i]/b[i],也就是比率,然后按这个从大到小排序,选最大的k个
但这是错的,一个很简单的反例:
3 1
100 1 100
100 5 200
正确答案应该是(101/105)*100 ≈ 96
#include<stdio.h> #include<algorithm> #include<math.h> using namespace std; typedef struct Res { double a, b; double d; bool operator < (const Res &b) const { if(d>b.d) return 1; return 0; } }Res; Res s[1005]; int main(void) { int i, n, k; double cnt, sa, sb; while(scanf("%d%d", &n, &k), n!=0 || k!=0) { k = n-k; for(i=1;i<=n;i++) scanf("%lf", &s[i].a); for(i=1;i<=n;i++) scanf("%lf", &s[i].b); cnt = 0; while(1) { for(i=1;i<=n;i++) s[i].d = s[i].a-cnt*s[i].b; sort(s+1, s+n+1); sa = sb = 0; for(i=1;i<=k;i++) { sa += s[i].a; sb += s[i].b; } if(fabs(sa/sb-cnt)<0.00001) break; cnt = sa/sb; } printf("%.0f\n", cnt*100); } return 0; }
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